Java编译时出现 No enclosing instance of type Main is accessi

今天在编译Java程序的时候出现以下错误： No enclosing instance of type Main is accessible. Must qualify the allocation with an enclosing instance of type Main (e.g. x.new A() where x is an instance of Main). 我原来编写的源代码是这样的： publ

今天在编译Java程序的时候出现以下错误：

No enclosing instance of type Main is accessible. Must qualify the allocation with an enclosing instance of type Main (e.g. x.new A() where x is an instance of Main).

我原来编写的源代码是这样的：

public class Main 
{
class Dog //定义一个“狗类”
{
private String name;
private int weight;
public Dog(String name, int weight) 
{
this.setName(name);
this.weight = weight;
}
public int getWeight() 
{
return weight;
}
public void setWeight(int weight) 
{this.weight = weight;}
public void setName(String name)
{this.name = name;}
public String getName() 
{return name;}
}
public static void main(String[] args)
{
Dog d1 = new Dog("dog1",1);

}
}

出现这个错误的时候，我一直不太理解。

在借鉴别人的解释之后才恍然大悟。

在代码中，我的Dog类是定义在Main中的内部类。Dog内部类是动态的内部类，而我的main方法是static静态的。

就好比静态的方法不能调用动态的方法一样。

有两种解决办法：

第一种：

将内部类Dog定义成静态static的类。

第二种：

将内部类Dog在Main类外边定义。

修改后的代码：

第一种：

public class Main 
{
	public static class Dog 
	{
		private String name;
		private int weight;
		public Dog(String name, int weight) 
		{
			this.setName(name);
			this.weight = weight;
		}
		public int getWeight() 
		{
			return weight;
		}
		public void setWeight(int weight) 
		{this.weight = weight;}
		public void setName(String name)
		{this.name = name;}
		public String getName() 
		{return name;}
	}
	public static void main(String[] args)
	{
		Dog d1 = new Dog("dog1",1);	
	}
}

第二种：

public class Main 
{
	public static void main(String[] args)
	{
		Dog d1 = new Dog("dog1",1);	
	}
}

class Dog 
{
		private String name;
		private int weight;
		public Dog(String name, int weight) 
		{
			this.setName(name);
			this.weight = weight;
		}
		public int getWeight() 
		{
			return weight;
		}
		public void setWeight(int weight) 
		{this.weight = weight;}
		public void setName(String name)
		{this.name = name;}
		public String getName() 
		{return name;}
}